首先，SCANF为下一个scanfs获得其他Chrachter

发布于 2025-02-09 18:28:59 字数 393 浏览 3 评论 0原文

我正在在C中制作一个简单的计算器，我希望该程序显示消息“输入操作员！”当用户输入不是操作员的东西时，请输入，空间或选项卡。但是，如果用户输入类似123的数字，则显示3次消息，而不是一次。我该如何解决？

char op;
printf("please choose an operator (+,-,*,/): ");
while (op!='+' && op!='*' &&op!='-' &&op!='/' )
    {
    if (op!='\n'&& op!= '\t'&& op!=' ')
    printf ("Enter an operator!\n");
    scanf ("%c",&op);
    }

原文

I'm making a simple calculator in C and I want the program to show the message "Enter an operator!" when the user enters something that isn't an operator, enter, space or tab. but if the user enters a number like 123 the message is shown 3 times instead of once. how can I fix this?

char op;
printf("please choose an operator (+,-,*,/): ");
while (op!='+' && op!='*' &&op!='-' &&op!='/' )
    {
    if (op!='\n'&& op!= '\t'&& op!=' ')
    printf ("Enter an operator!\n");
    scanf ("%c",&op);
    }

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还在原地等你 2025-02-16 18:28:59

可以使用扫描集％1 [+ - /*]将在一组运算符中扫描一个字符。
如果字符不在集合中，scanf将返回0。清洁输入流，然后重试。

#include <stdio.h>
#include <stdlib.h>

int main ( void) {
    char op[2] = "";
    int result = 1;

    printf ("Enter an operator!\n");
    do {
        if ( result != 1) {
            printf ("Enter an operator!\n");
        }
        result = scanf ( "%1[+-/*]", op);
        if ( result == EOF) {
            fprintf ( stderr, "EOF\n");
            return 1;
        }
        int clean = 0;
        while ( ( clean = fgetc ( stdin)) != '\n') { // clean input stream
            if ( clean == EOF) {
                fprintf ( stderr, "EOF\n");
                return 1;
            }
        }
    } while ( result != 1);

    printf ( "operator %s\n", op);

    return 0;
}

A scan set could be used %1[+-/*] will scan one character in the set of operators.
If the character is not in the set, scanf will return 0. Clean the input stream and try again.

#include <stdio.h>
#include <stdlib.h>

int main ( void) {
    char op[2] = "";
    int result = 1;

    printf ("Enter an operator!\n");
    do {
        if ( result != 1) {
            printf ("Enter an operator!\n");
        }
        result = scanf ( "%1[+-/*]", op);
        if ( result == EOF) {
            fprintf ( stderr, "EOF\n");
            return 1;
        }
        int clean = 0;
        while ( ( clean = fgetc ( stdin)) != '\n') { // clean input stream
            if ( clean == EOF) {
                fprintf ( stderr, "EOF\n");
                return 1;
            }
        }
    } while ( result != 1);

    printf ( "operator %s\n", op);

    return 0;
}

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木森分化 2025-02-16 18:28:59

首先，使用格式字符串“％c”输入字符。请注意格式字符串中的领先空间。它允许跳过白空间字符，例如新行字符'\ n'。

请注意，您正在使用非初始化的变量OP在WARE循环的条件下导致行为不确定。

char op;
printf("please choose an operator (+,-,*,/): ");
while (op!='+' && op!='*' &&op!='-' &&op!='/' )
//...

简化的方法可以看起来以下方式

#include <stdio.h>
#include <string.h>

//...

char op;
printf("please choose an operator (+,-,*,/): ");
const char *operations = "+*-/";

while ( 1 )
{
    int result = scanf( " %c", &op );

    if ( result == 1 && strchr( operations, op ) != NULL )
    {
        break;
    }

    if ( result == 1 )
    {
        scanf( "%*[^\n]" );
    } 

    printf ("Enter an operator!\n");
}

For starters use the format string " %c" to enter a character. Pay attention to the leading space in the format string. It allows to skip white space characters as for example the new line character '\n'.

Pay attention to that you are using an uninitialized variable op in the condition of the while loop that results in undefined behavior.

char op;
printf("please choose an operator (+,-,*,/): ");
while (op!='+' && op!='*' &&op!='-' &&op!='/' )
//...

A simplified approach can look the following way

#include <stdio.h>
#include <string.h>

//...

char op;
printf("please choose an operator (+,-,*,/): ");
const char *operations = "+*-/";

while ( 1 )
{
    int result = scanf( " %c", &op );

    if ( result == 1 && strchr( operations, op ) != NULL )
    {
        break;
    }

    if ( result == 1 )
    {
        scanf( "%*[^\n]" );
    } 

    printf ("Enter an operator!\n");
}

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